Question

An iceberg,having a volume of 2060 $0.92\frac{g}{c{m}^3}$,floats in sea water (density = 1.03$\frac{g}{c{m}^3}$ ) with a portion of 224 $c{m}^3$ above the surface. Calculate the density of ice.

Answer 1

Step 1:Given :

1. Density of water = ${\rho }_2$=1.03 $\frac{g}{c{m}^3}$.
2. Volume of ice berg = $V_1$= 2060 $c{m}^3$.
3. Volume of sea water above the surface of ice berg = $V_2=224c{m}^3$.
4. We have to find density of ice = ?
5. Volume of water displaced $V^1$= $V_1$-$V_2$.
6. Weight of iceberg =weight of water displaced ( $V{\rho }_{1}g=V^1{\rho }_{2}g$) .

Step 2:Calculation :

Volume of water displaced $V^1$= $V_1$-$V_2$.

Volume of water displaced V = (2060 - 224 ) $c{m}^3$.

V = 1836 $c{m}^3$.

Weight of iceberg =weight of water displaced ( $V{\rho }_{1}g=V^1{\rho }_{2}g$) .

2060$c{m}^3$$\times$${\rho }_1$$\times$g = 1836$c{m}^3$$\times$1.03$\frac{g}{c{m}^3}$$\times$g.

2060$c{m}^3$$\times$${\rho }_1$=1891.08 $g$.

${\rho }_1$=$\frac{1891.08g}{2060c{m}^3}$= 0.918 $\frac{g}{c{m}^3}$.

Density of water ${\rho }_1$=$0.92\frac{g}{c{m}^3}$.

Hence, Density of water ${\rho }_1$=$0.92\frac{g}{c{m}^3}$.